Question: $f(t) = 4t$ $h(t) = -t-3-f(t)$ $g(n) = 4n-4(h(n))$ $ g(h(2)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(2)$ . Then we'll know what to plug into the outer function. $h(2) = -2-3-f(2)$ To solve for the value of $h$ , we need to solve for the value of $f(2)$ $f(2) = (4)(2)$ $f(2) = 8$ That means $h(2) = -2-3-8$ $h(2) = -13$ Now we know that $h(2) = -13$ . Let's solve for $g(h(2))$ , which is $g(-13)$ $g(-13) = (4)(-13)-4(h(-13))$ To solve for the value of $g$ , we need to solve for the value of $h(-13)$ $h(-13) = -(-13)-3-f(-13)$ To solve for the value of $h$ , we need to solve for the value of $f(-13)$ $f(-13) = (4)(-13)$ $f(-13) = -52$ That means $h(-13) = -(-13)-3-(-52)$ $h(-13) = 62$ That means $g(-13) = (4)(-13)+(-4)(62)$ $g(-13) = -300$